A196730
Numbers m such that the sum of the distinct residues of x^m (mod m) is a perfect square, x=0..m-1.
Original entry on oeis.org
1, 2, 4, 8, 9, 10, 16, 26, 32, 34, 58, 64, 74, 81, 82, 84, 106, 122, 128, 146, 178, 194, 196, 202, 218, 226, 250, 256, 274, 298, 314, 346, 361, 362, 386, 394, 441, 458, 466, 480, 482, 512, 514, 538, 554, 562, 586, 626, 634, 674, 676, 698, 706, 722, 729, 746
Offset: 1
a(8) = 26 because x^26 == > 0, 1, 3, 4, 9, 10, 12, 13, 14, 16, 17, 22, 23, 25 (mod 26), and the sum = 169 = 13^2.
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sumSquares := proc(n)
local re, x, r ;
re := {} ;
for x from 0 to n-1 do
re := re union { modp(x^n, n) } ;
end do:
add(r, r=re) ;
end proc:
for n from 1 to 750 do
z:= sqrt(sumSquares(n));
if z=floor(z) then
printf("%d, ", n);
end if;
end do: #
A351177
Number of distinct residues of k^(n^2) (mod n^2+1), k=0..n^2.
Original entry on oeis.org
2, 2, 10, 2, 26, 2, 42, 8, 82, 2, 122, 16, 170, 2, 226, 2, 290, 12, 362, 2, 170, 50, 530, 2, 626, 2, 90, 80, 842, 70, 962, 36, 130, 92, 1226, 2, 1370, 138, 1522, 2, 1626, 178, 1554, 152, 2026, 152, 2210, 232, 2402, 12, 2602, 272, 2810, 2, 306, 2, 1010, 338, 3482
Offset: 1
a(2) = 2 because k^(2^2) == 0, 1 (mod 5) implies 2 distinct residues.
The table of k^(n^2) (mod n^2+1) of residues starts in row n=1 with columns k>=2 as:
0,1;
0,1,1,1,1;
0,1,2,3,4,5,6,7,8,9;
0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25;
Its row sums are 1, 4, 45, 16, 325, ...
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a:= n-> nops ({seq (k&^(n^2) mod (n^2+1), k=0..n^2)}):
seq (a(n), n=1..100);
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Table[Length[Union[PowerMod[Range[0,n^2],n^2,n^2+1]]],{n,100}]
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a(n) = #Set(vector(n^2+1, k, k--; Mod(k, n^2+1)^n^2)); \\ Michel Marcus, Mar 18 2022
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